MCQ Questions for Class 10 Maths Real Numbers Quiz 2

The number of possible pairs of number, whose product is 5400 and the HCF is 30 is

0%
1
0%
2
0%
3
0%
4

Explanation

$Given\quad that\quad product\quad of\quad the\quad number\quad is\quad 5400=30\times 3\times 2\times 30.\\ \therefore \quad Possible\quad pairs\quad as\quad per\quad the\quad requirment\quad are-\\ (1)\quad 30\times (3\times 2\times 30)=30\times 180\\ (2)\quad (30\times 3)\times (2\times 30)=90\times 60\\ \therefore \quad Total\quad number\quad of\quad pairs=2\quad \quad (Ans)$

$\dfrac{35}{50}$ has a non-terminating decimal expansion.

0%
True
0%
False

Explanation

Let us write the denominator of the given fraction as a product of its prime factors,

$50=2\times5\times5$

$=2\times5^2$

So, we can observe that the denominator of the given fraction is of the for

$2^m \times 5^n$ so, fraction

$\dfrac{35}{50}$ will have terminating decimal expansion.

Hence, the given statement is false.

According to Euclid's division algorithm, HCF of any two positive integers

$a$ and

$b$ with

$a > b$ is obtained by applying Euclid's division lemma to

$a$ and

$b$ to find

$q$ and

$r$ such that

$a = bq + r$ , where

$r$ must satisfy

0%
$1< r< b$
0%
$0< r< b$
0%
$0\leq r< b$
0%
$0< r\leq b$

Explanation

According to Euclid's division algorithm, HCF of any two positive integers

$a$ and

$b$ with

$a > b$ is obtained by applying Euclid's division lemma to

$a$ and

$b$ to find

$q$ and

$r$ such that

$a = bq + r$

The remainder

$r$ is either equal to or greater than

$0$ but it is always smaller than divisor

$b$ .

i.e

$0\le r<b$

Hence, option C is correct.

0%
Both Assertion and Reason are correct and Reason is the correct explanation for Assertion.
0%
Both Assertion and Reason are correct, but Reason is not the correct explanation for Assertion.
0%
Assertion is correct, but Reason is incorrect.
0%
Assertion is incorrect, but Reason is correct.

Explanation

In a fraction, if the denominator is of the form

$2^n\times5^m$ then the fraction is always terminating.
In this case,

$34.12345=\dfrac{3412345}{100000}$
Denominator is

$100000=10^5=2^5\times5^5$
Denominator is of the form

$2^n\times5^m$
So, it is terminating and reason is the correct explanation to assertion.
Option

$A$

The decimal representation of

$\dfrac {93}{1500}$ will be:

0%
terminating
0%
non-terminating
0%
non-terminating, repeating
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non-terminating, non-repeating

Explanation

$\dfrac {93}{1500} =\dfrac{3\times 31}{3\times 5\times 5\times 5\times 2\times 2 } =0.062$

Also, in fully reduced form,

$\dfrac {93}{1500}$ has a denominator composed of

$2$ 's and

$5$ 's only (

$500=2^2\times 5^3$ ).

Therefore, the decimal representation will be terminating.

Thus, option

$A$ is correct.

The rational number which can be expressed as a terminating decimal is:

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$\displaystyle \frac{1}{6}$
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$\displaystyle \frac{1}{12}$
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$\displaystyle \frac{1}{15}$
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$\displaystyle \frac{1}{20}$

Explanation

A terminating decimal is a decimal that ends after a finite number of steps. It's a decimal with a finite number of digits.

Also, a decimal is terminating if in fully reducible form, the denominator is composed of

$2$ 's and

$5$ 's only.

Here, only option

$D$ has a denominator composed of

$2$ 's and

$5$ 's, i.e.

$20=2^2\times5$

and

$\dfrac {1}{20}= 0.05$ .

In all the other options, the decimal does not end with a finite number of digits and the denominator has factors other than

$2$ 's and

$5$ 's.

Therefore, option

$D$ is correct.

State True or False.

$-\displaystyle\frac{2}{5}\sqrt{8}$ is an irrational number.

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True
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False

Explanation

$\\ { \frac { -2 }{ 5 } \sqrt { 8 } }={ \frac { -4 }{ 5 } \sqrt { 2 } }=-0.8*\sqrt { 2 } \\ \sqrt { 2 } =1.41421356237........\\ \\ \sqrt { 2 } is\quad an\quad irrational\quad number,\quad since\quad its\quad decimal\quad representaion\quad is\quad non\quad terminating\quad non\quad repeating.\\ Multiplication\quad of\quad rational\quad with\quad irrational\quad is\quad irrational.\\ Hence,\quad { (\frac { -2 }{ 5 } \sqrt { 8 } ) }\quad is\quad an\quad irrational\quad number.\\ \quad$

The HCF of

$256,442$ and

$940$ is

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$2$
0%
$14$
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$142$
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None of these

Explanation

Prime factors of numbers are

$256=2\times 2\times 2\times 2\times 2\times 2\times 2\times 2\\ 442=2\times 13\times 17\\ 940=2\times 2\times 5\times 47$

HCF is the highest common factor among the numbers.

Thus among the given numbers

$2$ is the common factor. Hence HCF of given numbers is

$2$ .

So, correct answer is option A.

Which of the following are not rational?

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$\dfrac{2}{\sqrt{3}}$
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$\sqrt{16}$
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$\pi$
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$2 + \sqrt{25}$

Explanation

Only B and D are rational.
In A,

$\sqrt{3}$ is not an Integer. hence,

$\dfrac{2}{\sqrt{3}}$ it is not rational
In C,

$\pi = 3.142..$ is a non terminating non recurring decimal. Hence it is not rational.

Classify the following numbers as rational or irrational :

$2-\sqrt{5}$

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Irrational number
0%
Rational number
0%
Less Data
0%
None of the above

Explanation

$2$ is rational

$\sqrt 5 =2.035.........$ which is non terminating and non repeating hence irrational number.

We know that,

$\text{rational}$

$-$

$\text{irrational}= \text{irrational number.}$

Hence

$2-\sqrt 5= irrational \, number$

Hence, option A is the correct answer.

State TRUE or FALSE

$(2+\sqrt{3})(2-\sqrt{3})$ is Irrational

0%
True
0%
False

Explanation

we know that,

$a^2-b^2$ =

$(a+b)$

$(a-b)$

${ (2+\sqrt { 3 } ) }({ 2-\sqrt { 3 } ) }=4-3=1\\ Hnece\quad rational.\\$

State TRUE or FALSE

${(2+\sqrt{3})}^{2}$ is Irrational

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True
0%
False

Explanation

${ (2+\sqrt { 3 } ) }^{ 2 }=4+3+4\sqrt { 3 } =7+4\sqrt { 3 } \\ The\quad above\quad given\quad expression\quad consists\quad of\quad an\quad algebric\quad equation\quad \quad \\ consisting\quad of\quad irrational\quad terms,\quad hence\quad it\quad is\quad an\quad irrational\quad expression.\\$

Every irrational number is:

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a surd
0%
a prime number
0%
not a surd
0%
none

Explanation

An irrational number is a real number that cannot be represented as a ratio or a simple fraction.

By definition, a surd is an irrational root of a rational number. So we know that surds are always irrational and they are always roots.

For eg,

$\sqrt2$ is a surd since 2 is rational and

$\sqrt 2$ is irrational.

Similarly, the cube root of 9 is also a surd since 9 is rational and the cube root of 9 is irrational.

On the other hand,

$\sqrtπ$ is not a surd even though

$\sqrtπ$ is irrational because π is not rational.

Thus, to answer the question, every surd is an irrational number, though an irrational number may or may not be a surd

The answer is Option C.

HCF of the two numbers =

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Product of numbers + their LCM
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Product of numbers - their LCM
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Product of numbers $\times$ their LCM
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Product of numbers $\div$ their LCM
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Answer required

Explanation

The product of highest common factor

$(H.C.F.)$ and lowest common multiple

$(L.C.M.)$ of two numbers is equal to the product of two numbers.

If two numbers are

$a$ and

$b$ then

$HCF\ \times LCM=a\times b$

Hence,

$HCF=\dfrac{ab}{LCM}$

For example:

Let

$a=10\Rightarrow 2\times 5$ and

$b=15\Rightarrow 3\times 5$

So, the

$LCM$ of both numbers

$=2\times 3\times 5\Rightarrow 30$

Then

$HCF=\dfrac{10\times 15}{30}\Rightarrow 5$

Hence,

$HCF=\dfrac{Product\ of\ numbers}{Their\ LCM}.$

$20$ is written as the product of primes as :

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${2\times 5 }$
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${2\times 2\times 3\times 5}$
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${2\times 2\times 5}$
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${2\times 2\times 3}$

Explanation

To write a number as product of its primes, we divide it by various prime numbers $2, 3, 5, 7$ etc one by one and check by which prime numbers it is divisible with and how many times.

Hence, $20 = 2 \times 10 = 2 \times 2 \times 5$

Without actually performing the long division, state whether the following rational number will have a terminating decimal expansion or non -terminating decimal expansion

$\displaystyle \frac{7}{210}$

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Terminating decimal expansion
0%
Non -terminating decimal expansion
0%
Cannot be determined
0%
None

Explanation

Simplify it by dividing nominator and denominator both by

$7$ we get

$\dfrac{1}{30}$
Factorize the denominator we get

$30=2 \times 3 \times 5$
Denominator has

$3$ also in denominator
So denominator is not in form of

$2^{n} \times 5^{n}$
Hence it is non terminating.

Prime factors of

$140$ are :

0%
${2\times2\times 7}$
0%
${2\times2\times5}$
0%
${2\times2\times5\times7}$
0%
${2\times2\times5\times7\times3}$

Explanation

Factors of

$140$ are

$1,140,7,20,2,70,4,35,5, 28,10,14$

Prime factors are

$2, 7$ and

$5.$

So,

$140$ written as product of primes is

$2\times 2 \times 5\times 7.$

Euclids division lemma, the general equation can be represented as .......

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$a = b\times q + r$
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$a = b\div q + r$
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$a = b- q - r$
0%
$a = b + q - r$

Explanation

Euclids division lemma states that for any two positive integers

$a$ and

$b$ we can find two whole numbers

$q$ and

$r$ such that

$a = b \times q + r$ where

$0 \leq r < b.$
Therefore,

$A$ is the correct answer.

The statement dividend

$=$ divisor

$\times$ quotient

$+$ remainder is called

0%
Euclid's multiplication Lemma
0%
Euclid's addition Lemma
0%
Euclid's subtraction Lemma
0%
Euclid's division Lemma

Explanation

Euclids division lemma states that

$a = b \times q + r$ where

$0 \leq r < b.$

$a =$ dividend

$,b =$ divisor

$, q =$ quotient and

$r =$ remainder
Therefore,

$D$ is the correct answer.

$(3 + \sqrt {5})$ is ..............

0%
whole number
0%
an integer
0%
rational
0%
irrational

Explanation

Sum or difference of a rational and irrational number is irrational. Therefore,

$D$ is the correct answer.

$2\times 2\times 2\times 3\times 3\times 13 = 2^{3} \times 3^{2} \times 13$ is equal to

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$1004$
0%
$828$
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$724$
0%
$936$

Explanation

$936 = 2\times 2\times 2\times 3\times 3\times 13 = 2^{3} \times 3^{2} \times 13$
Therefore

$, D$ is the correct answer.

Which option will have a terminating decimal expansion?

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$\dfrac {77}{210}$
0%
$\dfrac {23}{30}$
0%
$\dfrac {125}{441}$
0%
$\dfrac {23}{8}$

Explanation

The decimal expansion of a number is terminating if the denominator of the number can be expressed in the form of $2^m\times 5^n$ where $m$ and $n$ are non-negative integers

A) $\dfrac{77}{210},$ prime factorisation of the denominator is given as

$210=2\times 3\times 5\times 7$

$\implies$ denominator is not of the form $2^m \times 5^n$

B) $\dfrac{23}{30},$ prime factorisation of the denominator is given as

$30=2\times 3\times 5$

$\implies$ denominator is not of the form $2^m \times 5^n$

C) $\dfrac{125}{441},$ prime factorisation of the denominator is given as

$441=21^2$

$\implies$ denominator is not of the form $2^m \times 5^n$

D) $\dfrac{23}{8},$ prime factorisation of the denominator is given as

$8=2^3$

$\implies$ denominator is of the form $2^m \times 5^n$ where $m=3$ and $n=0$

Hence the decimal expansion of $\dfrac{23}{8}$ is terminating

Fundamental theorem of arithmetic is also called as ______ Factorization Theorem.

0%
Algebra
0%
Ambiguous
0%
Unique
0%
None of these

Explanation

$30 = 2 \times 3 \times 5$ , where

$2$ and

$3$ are prime numbers.

We cannot get the number

$30$ by multiplying any other prime numbers.

Example :

$30 \neq 2 \times 5 \times 7.$

It is only with one particular set of prime numbers.

Hence, it is called Unique Factorization Theorem.

Therefore,

$C$ is the correct answer.

We need blocks to build a building. In the same way _______ are the basic blocks to form all natural numbers.

0%
prime numbers
0%
real numbers
0%
unique numbers
0%
negative numbers

Explanation

Let us consider some examples.

$10 = 2 \times 5$ , we need

$2$ and

$5$ which are prime numbers to get a new number

$10$ .

$12= 2 \times 2 \times 3$ , we need

$2$ and

$3$ which are prime numbers to get a new number

$12$ .
Therefore

$,$

$A$ is the correct answer.

Using fundamental theorem of Arithmetic find L.C.M. and H.C.F of

$816$ and

$170$ .

0%
L.C.M. $=4080$ and H.C.F $=34$
0%
L.C.M. $=8160$ and H.C.F $=68$
0%
L.C.M. $=2048$ and H.C.F $=62$
0%
L.C.M. $=2040$ and H.C.F $=72$

Explanation

According to the fundamental theorem of arithmetic every composite number can be factorised as a product of primes and this factorization is unique apart from the order in which the prime factor occurs.

Fundamental theorem of arithmetic is also called unique factorization theorem.
Composite number = product of prime numbers.
Any Integer greater than 1, either be a prime number or can be written as a product of prime factors.

The prime factors of

$816=2\times2\times2\times2\times3\times17=2^4\times3\times17$

The prime factors of

$170=2\times5\times17$

LCM of

$816$ and

$170=2^4\times3\times5\times17=4080$

HCF of

$816$ and

$170=2\times17=34$

Option A.

State whether the given statement is True or False :

$2\sqrt { 3 }-1$ is an irrational number.

0%
True
0%
False

Explanation

$\text{Here 1 is a rational number and }$

$2\sqrt3$ is a

$irrational$ number

And

$\text{the difference of rational and irrational is always an irrational number}$

So that

$(2\sqrt3- 1)$ is an irrational number.

hence option A is correct.

State true or false

$\sqrt{2} + \dfrac{1}{\sqrt{2}}$ is irrational.

0%
True
0%
False

Explanation

$\sqrt2\rightarrow irrational\ number\\\dfrac1 {\sqrt2}=\dfrac{\sqrt2}{2}\rightarrow irrational\ number$

Sum of two irrational numbers is an irrational number. Thus given number is an irrational number.

Without actually dividing find which of the following are terminating decimals.

0%
$\dfrac{3}{25}$
0%
$\dfrac{11}{18}$
0%
$\dfrac{13}{20}$
0%
$\dfrac{41}{42}$

Explanation

We know that

$\cfrac{p}{q}$ is terminating if p and q are co-prime and q is of the form

${2}^{m} {5}^{n}$ ; where m and n are non-negative integers.

(A)

$\cfrac{3}{25}$ :

Checking co-prime-

Since 3 and 25 have no common factors, hence 3 and 25 are co-prime.

Now,

$25 = 5 \times 5 = {5}^{2}$

$\therefore \text{ Denominator} = {5}^{2} = 1 \times {5}^{2} = {2}^{0} \times {5}^{2}$

$\therefore$ Denominator is of the form

${2}^{m}{5}^{n}$ , where m = 0 and n = 2.

Hence,

$\cfrac{3}{25}$ is a terminating decimal.

(B)

$\cfrac{11}{18}$ :

Checking co-prime-

Since 11 and 18 have no common factors, hence 11 and 18 are co-prime.

Now,

$18 = 2 \times 3 \times 3 = 2 \times {3}^{2}$

$\therefore$ Denominator is not of the form

${2}^{m}{5}^{n}$ .

Hence,

$\cfrac{11}{18}$ is not a terminating decimal.

(C)

$\cfrac{13}{20}$ :

Checking co-prime-

Since 13 and 20 have no common factors, hence 13 and 20 are co-prime.

Now,

$20 = 2 \times 2 \times 5 = {2}^{2} \times 5$

$\therefore \text{ Denominator} = {2}^{2} \times 5 = {2}^{2} \times {5}^{1}$

$\therefore$ Denominator is of the form

${2}^{m}{5}^{n}$ , where m = 2 and n = 1.

Hence,

$\cfrac{13}{20}$ is a terminating decimal.

(D)

$\cfrac{41}{42}$ :

Checking co-prime-

Since 41 and 42 have no common factors, hence 41 and 42 are co-prime.

Now,

$42 = 2 \times 3 \times 7$

$\therefore$ Denominator is not of the form

${2}^{m}{5}^{n}$ .

Hence,

$\cfrac{41}{42}$ is not a terminating decimal.

$\sqrt { 7 }$ is a

0%
an integer
0%
an irrational number
0%
a rational number
0%
none of these

Explanation

$\sqrt 7$ is a irrational number

State True or False:

$4\, - \,5\sqrt 2$ is irrational if

$\sqrt 2$ is irrational.

0%
True
0%
False

Explanation

The given number is

$4-5\sqrt{2}$

$\sqrt{2}$ is irrational then

$5\sqrt{2}$ is also irrational because product of a rational and a irrational number is also irrational.

Again, we know, the difference of a rational and a irrational number is also irrational.

So,

$4-5\sqrt{2}$ is irrational.

Hence, the given statement is

$True$ .

CBSE Questions for Class 10 Maths Real Numbers Quiz 2 - MCQExams.com

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Practice Class 10 Maths Quiz Questions and Answers

CBSE Questions for Class 10 Maths Real Numbers Quiz 2 - MCQExams.com

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Practice Class 10 Maths Quiz Questions and Answers

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